What is expand (81(4n^2+2n+1))/(4n^2) ?

The solution to expand (81(4n^2+2n+1))/(4n^2) is 81+(81)/(2n)+(81)/(4n^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand (81(4n^2+2n+1))/(4n^2)

expand

\frac{81 ( 4 n ^{2} + 2 n + 1 )}{4 n ^{2}}

81 + \frac{81}{2 n} + \frac{81}{4 n ^{2}}

Solution steps

\frac{81 ( 4 n ^{2} + 2 n + 1 )}{4 n ^{2}}

Expand

81 (4 n^{2} + 2 n + 1) : 324 n^{2} + 162 n + 81

= \frac{324 n ^{2} + 162 n + 81}{4 n ^{2}}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{324 n ^{2} + 162 n + 81}{4 n ^{2}} = \frac{324 n ^{2}}{4 n ^{2}} + \frac{162 n}{4 n ^{2}} + \frac{81}{4 n ^{2}}

= \frac{324 n ^{2}}{4 n ^{2}} + \frac{162 n}{4 n ^{2}} + \frac{81}{4 n ^{2}}

Cancel

\frac{324 n ^{2}}{4 n ^{2}} : 81

= 81 + \frac{162 n}{4 n ^{2}} + \frac{81}{4 n ^{2}}

Cancel

\frac{162 n}{4 n ^{2}} : \frac{81}{2 n}

= 81 + \frac{81}{2 n} + \frac{81}{4 n ^{2}}

e^{(1 + i) x}

u (6, - 4, - 2) v (5, 3, 4)

(vx - 2 x) (v d x + x d v)

e^{\frac{1}{2} (1 - x^{2})}

\frac{( x + 7 ) ( x - 4 )}{x - 2}

What is expand (81(4n^2+2n+1))/(4n^2) ?
The solution to expand (81(4n^2+2n+1))/(4n^2) is 81+(81)/(2n)+(81)/(4n^2)